(expires 2/18)     Fit the points $(-1.9,-4.7), (-0.8,1.2),
(0.1,2.8), (1.4,-1.2), (1.8,-3.5)$ by means of a quadratic function $f(x)=ax^2+bx+c$, using the least square method. First, do this step by step, as we did in class; then, use the built-in Maple command, described in the notes. Check that the two solutions agree.

(expires 2/18)     Fit the set of points

\begin{displaymath}(1.02,-4.30), (1.00,-2.12), (0.99,0.52), (1.03,2.51), (1.00,3.34),

with a line, using the least square method we used in class. You will see that this is not a good fit. Think of a better way to do the fit and use Maple to do it. Explain in your solution why you think your better way is better.

(expires 2/18)     In this problem we will estimate the charge of the electron: If an electron of energy $E$ is thrown into a magnetic field $B$, perpendicular to its velocity, its trajectory will be deflected into a circular trajectory of radius $r$. The relation between these three quantities is:
B\, r\, e = \frac{E^2}{m^2 c^4} \, \sqrt{E^2 - m^2 c^4},
\end{displaymath} (1)

where $e$ and $m$ are, respectively, the charge and the mass of the electron, and $c$ is the speed of light. The rest mass of the electron is defined as $E_0=mc^2$, and is about equal to $8.817 \: 10^{-14}$ Joules. In our experimental set-up the energy of the emitted electrons is set to be $E = 2.511 E_0$.
Use read to make Maple load and execute the commands in the file electron_data.txt, which is located in the Worksheets directory of the mat331 account. This defines a list called electron. Each element of the list is a pair of the form $[B_i,r_i]$, and these quantities are expressed in Teslas and meters. Use least square fitting to determine the best value for $e$. [Hint: Notice that the right hand side of (1) is just a constant--calculate it once and for all and give it a name. Then (1) is a very easy equation, which is linear in the unknown parameter $e$. To verify your solution: $e \approx 1.602 \:
10^{-19}$ Coulomb].
Physical constants courtesy of N.I.S.T.

(expires 2/18)     Prove relation (1), knowing the following physical facts: In relativistic dynamics Newton's law is replaced by
F = m \, \frac{d}{dt} \left( \frac{v} {\sqrt{1 -
\frac{v^2}{c^2}}} \right),
\end{displaymath} (2)

where $F$ is the force acting on a particle, $m$ its mass and $v$, a function of time, its velocity. In the case at hand, the force exerted by a magnetic field $B$ on an electron is $F = e\, v\, B / c$. Recall that in a circular motion the acceleration $a= dv/dt = v^2/r$, $r$ being the radius of the circle. Since (1) is expressed in terms of the energy, rather than the velocity, you also need Einstein's formula,
E = \frac{mc^2} {\sqrt{1 - \frac{v^2}{c^2}}},
\end{displaymath} (3)

which can be solved in terms of $v$.

MAT 331